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C. From statement 1-> K = \(2^a*5^b*7^c*P^z\) ; for any prime number P. There exist ∞ numbers with the same set of prime numbers. Hence not sufficient. From statement 2-> K<100. K can be any integer <100. Clearly not sufficient Combing them -> Only Value of K possible is 70.

14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p? A)2 B)3 C)5 D)10 E)Cannot be determined.

E. From statement 1-> p=2 or 2*3^2 or 2*2^2 or 2* Any-prime^even--> not sufficient From statement 2-> p=2 or 4 or any even <15 => Not sufficient Combining them -> p=2 or p=2*2^2=8 Hence p can be 2 or 8 Thus not sufficient

Hence E.

16)Data Sufficiency->What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is an even number less than 15.

C. From statement 1-> p=3 or 3*3^2 or 3*7^2 or 3*prime^even--> not sufficient From statement 2-> p=2 or 4 or any even <15 => Not sufficient Combining them -> p=3*2^2 =12 is the only possible value. Hence Sufficient. Hence C

17)10^25 – 560 is divisible by all of the following EXCEPT:- A)11 B)8 C)5 D)4 E)3

E. Clearly it is divisible by 8,5,4 The real Test here is choosing between 3 and 11. 10^t for any positive integer t will always leave a remainder 1 with 3. e.g-> 10=> 3*3+1 100=3*33+1 1000=3*333+1 10000=3*3333+1

Hence 10^25=3m+1 Also 560=3k+2 (sum of digits =11,hence remainder with 3-> 2) Thus 3m+1-3k-2 => 3x-1 for some integer x Clearly it will never be divisible by 3.

Alternatively,we can also use the "Binomial Expansion".

18)Data Sufficiency->What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is a factor of 75

B. As x and y are prime numbers both greater than 2 => they both must be odd Hence x+y=odd+odd=even x*y=odd*odd=odd Notice that if x=y => x/y will be 1. Hence III is not always true. If x and y would happen to be different primes then III will be always true.

22)Which of the following statements must be true-> A)zero is neither prime nor composite. B)one is neither prime nor composite. C)-3 and 3 are both primes.

A and B. Notice -3 is negative and hence cannot be prime

23)Which of the following statements must be true-> 1)A prime number must be positive. 2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p. 3)The product of first ten primes is even. 4)All prime numbers greater than 71 are odd. 5)2 and 3 are the only consecutive integers that are also prime numbers. 6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one. 7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n-1. 8)All the prime numbers greater than 3 can be written as either 6+1 or 6n-1.

A. Statement 1->Using the property => "Every number is a factor of itself " B must be a factor of B Hence B must be a factor of A. Thus A/B must be an integer -> Sufficient Statement 2-> We can use test cases to Prove this insufficient Case 1->\(\frac{2*3*7}{2*3*7}\) =1=Integer Case 2->\(\frac{2*3*7*11}{2^5}\)=> Not an integer. Hence insufficient Hence A. Source-> GMAT-Prep

26)Data Sufficiency->Is \(\frac{n}{14}\) an integer?

D. Statement 1->n=28k = 14*2k=14k' Statement 2->n=70k= 14*5k=14k' Hence D. Alternatively,we can use the property => "A number is divisible by factors as well as factors of its factors"

27)If x is a prime number, what is the value of x?

(1) There are a total of 50 prime numbers between 71 and x, inclusive. (2) There is no integer n such that x is divisible by n and 1 < n < x.

A. In statement 1->Notice that 71 is 20th prime. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71. Hence x will be 50th prime from 71. Hence sufficient. Statement 2-> This is just another way of saying that x is prime. Hence not sufficient Hence A.

28)Data Sufficiency->If p = (n)(5^x)(3^k), is p divisible by 10?

C. Firstly we are not given in p,x,k,n are an integers or not. If p is an integer -> Then for p to be divisible by 10 => It must be dividable by 5 and 2=> It must have both 5 and 2 as its primes. Hence => n must be even and n,k,x must be all pastime integers. Statement 1-> All n,k,x are even. Case 1-> \(-2*5^{-10}*3^{-10}\)=>p is not an integer Case 2->\(2*5^2*3^4\)=> P is an integer and is dividable by 10. Hence not sufficient. Statement 2-> All are positive Again lest use test cases Case 1->\(1*5*3\)=> p is not divisible by 10 Case 2-> \(2*5*3\)=> p is divisible by 10 Combing the two statements => n must be a positive integer and x,k are positive even numbers. Hence p must be dividable both 2 and 5. Hence p must be dividable by 10. Hence C.

29)Data Sufficiency->If x is a prime number,what is the value of x?

1)There are 3 prime numbers between 11 and x 2 )x+1 is a prime number

33)Data Sufficiency->If z is a positive integer,is (z^31+7)^2 divisible by 4? A)√z has five prime factors. B)All prime factors of z^3 are greater than 7.

B. Property in action -> For any number to be odd -> Two must not be its prime. Also-> \(X\) and \(X^n\) always have the exact same prime factors.

34)If x is a positive odd integer and y is a negative even integer, which of the following must be true?

A. x^3 + y is a positive odd integer B. x^2 + y^2 is a negative odd integer C. x^0 + y^11 is a negative odd integer D. x + y is a positive odd integer E. x + y is a negative odd integer

C.This question is based on a very important property -> \(X\) and \(X^n\) always have the exact same prime factors. So \(x^{37}\) will have the exact same prime factors as x. Thus,here the question is asking us to get the number of prime factors of \(x\). Statement 1-> \(16x\) has one more prime factor than \(x^4\) Now \(x\) and \(x^4\) will have the exact same prime factors. Hence \(16x\) has one more prime than \(x\) \(16x=2^4*x\) => Hence 2 cannot be the prime factor of \(x\). But we still don't have the number of prime factors of \(x\). Hence not sufficient. Statement 2-> \(2*x^{16}\) has two prime factors Here again \(x^{16}\) will have the exact same primes as \(x\). So as \(2*x^{16}\) has two prime factors => \(x\) can have either one prime factor if 2 isn't one of its prime . Or two prime factors if 2 is one of its prime. Hence not sufficient. Combining the two statements=> As 2 cannot be one of the primes of x => x must have only one prime factor Hence sufficient. Hence C.

38)If a positive integer x has 3 prime factors,how many prime factors does x^3 have? A)27 B)6 C)3 D)1 E)cannot be determined

C. \(14k=2*7*k\) To get the number of prime factors of 14k => we need the number of prime factors of k Statement 1-> Using the property -> \(X\) and \(X^n\) always have the exact same primes => k must have both 2 and 5 as its primes. General expression for k= \(2^a*5^b*P^z\) Where a and b are non negative integers and P is any other prime number and z≥0 Hence not sufficient. Statement 2-> Form this statement => k can be \(2^a\) or \(5^b\) or \(2^a*5^b\) Or k=1 Hence not sufficient. Combing the two statements => k must have only 2 and 5 as its prime factors Hence 14k will have 3 prime factors Hence Sufficient Hence C.

40)Data Sufficiency->If x is an integer, is x^3 divisible by 9?

(1) x^2 is divisible by 9. (2) x^4 is divisible by 9.

B. The question is just asking us the prime factors for n. Statement 1-> From this statement we can conclude that n can have either 4 or 3 prime factors. Hence Not sufficient Statement 2-> Using the Property => X and X^n always have the exact same prime factors => As n^2 has 4 prime factors => n myst have 4 prime factors too. Hence B.

42)If x and y are positive integers, are they consecutive?

E. Using test cases can be extremely helpful here. Clearly statement 1 and 2 are not sufficient on their own. Combining them lets make test cases => Case 1=> x=3 y=3 x^2*y^4 will be divisible by 9. Case 2=> x=18 y=1/2 x^2*y^4 will not be divisible by 9. Hence E.

45)Data Sufficiency->For positive integers x and y, x^2 = 350y. Is y divisible by 28? (1) x is divisible by 4. (2) x^2 is divisible by 28.

A. We are told that x and y are both positive integers. x^2=350y As x is an integer => x^2 must be a perfect square. x^2=350y=2*5^2*7*y Hence the general expression for y would be -> 2*7*p^z where p is an integer and z is even. We are asked if y is divisible by 28 or not. Clearly y is already divisible by 7 and 2 Hence for y to be divisible by 28=> p must be even too. Statement 1-> x is divisible by 4 Hence x^2 must have 2^4. As x=2*5^2*7*y => For x^2 to have a 2^4 => y must have a 2^3. Hence y must have a 2^3 and it already has a 7. Thus y must be divisible by 28 Hence sufficient. Statement2-> Here lets use test cases y=2*7 =>y will not be divisible by 28 y=2*7*2^2 => y will be divisible by 28. Hence not sufficient. Hence A.

46)Data Sufficiency-> How many Prime factors does positive integer \(n\) have ? A)\(n^3\) has 7 prime factors. B)\(√n\) has 7 prime factors.

D. \(X\) and \(X^n\) always have exact same prime factors. In other words => "Exponentiation does not change the prime factors of any positive integer X."

47)If x and y represent the number of factors of 90 and 147 respectively.What is the value of x-y?

C. 21600=2^5*3^3*5^2 For even factors -> The power of 2 should at-least be 1. Hence possible cases -> 5*4*3=>60. Additionally, for odd factors -> The power of 2 must be zero. Possible cases -> 1*4*3=> 12.

50)How many positive odd divisors does 540 have? A)6 B)8 C)12 D)15 E)24

B. 330=2*5*3*11 For Odd factors => Power of 2 must be zero Possible cases -> 1*2*2*2=>8 So 330 has 8 odd divisors including one. Excluding one => 330 must have 7 odd divisors. Hence B.

54)If \(N=2^7*3^5*5^6*7^8\). How many factors of \(N\) are divisible by 50 but NOT by 100? A)240 B)345 C)270 D)120 E)None of these

C. For factors to be divisible by 50 and not divisible by 100 --> Allowed powers of 2->one Allowed powers of 5->5 Allowed powers of 3->6 Allowed powers of 7->9 Hence possible cases ->1*5*6*9=>270. Hence C.

55)Data Sufficiency->How many positive factors does the positive integer x have?

(1) \(x\) is the product of 3 distinct prime numbers.

(2) \(x\) and \(3^7\) have the same number of positive factors.

C. From statement 1-> Number of factors of x will be (n+1) Since we don't have any clue of n => not sufficient. From statement 2-> n must be 2 as no other value of n will satisfy the equation. Since we don't have any clue of x => not sufficient. Combining the two statements => Number of factors of x-> n+1=2+1=3 Hence C.

57)Data Sufficiency->If 2,3,5 are the only prime factors of a positive integer p,what is the value of p? A)p>100 B)p has exactly 12 factors including 1 and 12.

C. Since 2,3,5 are the only prime factors of p=> The general expression for p will be \(2^a*3^b*5^c\) where a,b,c are positive integers. Statement 1->p>100. There exist infinite such cases. In-fact there are only a few cases where p will be less than 100. Hence not sufficient. Statement 2-> Factors =12 Hence (a+1)*(b+1)*(c+1)=12=2*2*3 NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers. Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1) Posible values of p=> 150,90,60 Hence not sufficient. Combining the two statements => p must be 150. Hence C.

This is probably the hardest Question in this set.

58)Data Sufficiency->Is the positive integer P prime? (1) 61<P<67 (2) P is not divisible by 2.

59)Which of the following statements is/are true? A)A perfect square always has an odd number of factors. B)If a given integer has an odd number of factors then that integer must be a perfect square.

Both A and B are true. Two Important Properties -> A perfect Square always has an odd number of factors If a number has an odd number of factors => It must be a perfect square.

60)How many factors does 64 have? A)8 B)6 C)7 D)6 E)4

C. Method 1=> 64=2^6 Hence it will have 7 factors Hence C. Method 2=> 64 is a perfect square =>Thus, It will always have an of Odd number of factors. The only answer choice that is odd is C.

61)Data Sufficiency->Is the positive integer \(n\) divisible by \(6\)? (1) \(\frac{n^2}{180}\) is an integer.

A. For n to be divisible by 6 => it must be divisible by both 2 and 3. Statement 1-> As n^2 is divisible by 180 n^2 must have both 2 and 3 as its prime factors. But X and X^p always have the exact same prime factors. Hence n must also have both 2 and 3 as its primes too. Hence n must be divisible by 6=> Sufficient Statement 2-> Using test cases -> n=1=> \(\frac{144}{n^2}\) will be an integer=>NO n is not divisible by 6. n=6=>\(\frac{144}{n^2}\) will be an integer => YES n is divisible by 6. Hence not Sufficient. Hence A.

62)If a positive integer x has 5 prime factors, how many prime factors does 3x have? A) 5 B) 6 C) 8 D) 15 E) cannot be determined

E. Let \(x=(p1)^a *(p2)^b *(p3)^c *(p4)^d *(p5)^e\) Where \(p1,p2,p3,p4,p5\) are prime factors of x and a,b,c,d,e are positive integers. \(3x=3*(p1)^a *(p2)^b *(p3)^c *(p4)^d *(p5)^e\) Hence if any one of p1,p2,p3,p4,p5 is 3 => 3x will have 5 prime factors. Else => 3x will have 6 prime factors. Hence 3x may have 5 or 6 prime factors. So there is no unique answer. Hence E.

63)How many positive distinct prime factors does 5^20 + 5^17 have? A) One B) Two C) Three D) Four E) Five

B. Property->In order check whether a given number is prime or not => We just need to divide it by all the prime numbers less than or equal to the square root of that number. Highest value in this set-> 220 \(√220\)=14.something Hence we need to check the divisibility with primes less than 14.something => {2,3,5,7,11,13} All primes >2 are odd So to shorten our calculation we just need to check the odd numbers. 201-> Divisible by 3 203->Divisible by 7 205->Divisible by 5 207->Divisible by 3 209->Divisible by 11 211-> PRIME=> Not divisible by any prime numbers in our list. 213-> Divisible by 3 215->Divisible by 5 217->Divisible by 7 219->Divisible by 3 Hence only 211 is a prime number. Hence B.

65)Data Sufficiency->If y is a positive odd integer less than 33,does y have a factor p such that 1<p<y? A)y+2 is divisible by 3. B)Units digit of y is 7.

C. Note-> Does y have a factor p such that 1<p<y can be rephrased as -> Is y prime? The Question is just asking us if y prime or not. Statement 1-> y=> {1,4,7,10,13,16,19,22,25,28,31} Hence not sufficient. Statement 2-> y can be -> 7-> Prime 17-> Prime 27-> Composite Hence not sufficient Combing the two statements -> y=7 is the only possible value that satisfies both the statements. Hence y=7 So y is prime Hence C.

66)If x is a positive integer less than 100, for how many values of x is x/6 a prime number? A)2 B)6 C)8 D)13 E)17

67)Data Sufficiency->If n is an integer between 10 and 99, is n < 80? (1) The sum of the two digits of n is a prime number. (2) Each of the two digits of n is a prime number.

B. We can use test cases to arrive at an answer here. Statement 1-> n=92=> sum=11 which is prime number. So n is not less than 80. n=12=>sum=3 which is a prime number. So n is less than 80. Hence not sufficient. Statement 2-> Both the digits of n are prime. Allowed tens digit of n->{3,5,7} So n must be of the form 3X or 5X or 7X where X is the units digit of n. So,n can never be greater than 80 as both 8 and 9 are non prime numbers. Hence sufficient. Hence B. Source->Gmat-Prep

68)If n is a positive integer, which of the following statements are correct? A)All the prime numbers greater than 3 can be represented in the form 4n +1 or 4n + 3. B)All numbers of the form 4n + 1 and 4n + 3 are prime numbers. C)All the prime numbers greater than 3 can be represented in the form 6n – 1 or 6n + 1. D)All numbers of the form 6n + 1 and 6n - 1 are prime numbers.

A and C are correct. An important property of prime numbers ->Every prime number greater than 3 can be written as=> 6n+1 OR 6n-1 6n+5 OR 6n-5 4n+1 OR 4n-1 4n+3 OR 4n-3 This can be verified taking the first few primes. But alternatively every number of these forms isn't prime. E.g-> 25=6n+1 or 4n+1 and it isn't prime.

69)Data Sufficiency->If p is an integer, is p + 4 a prime number less than 50? (1) p + 1 is a prime number. (2) p is a prime number.

A. We are told that p is an integer greater than 1 and we are asked if p is a prime number or not. Statement 1-> p is a factor of 13. But 13 has got only 2 factors as it is a prime number. Factors => 1 and 13 so p must be either 1 or 13. But p>1 Hence p must be 13. So p is definitely a prime number. Sufficient. Statement 2-> p is factor of 78. Lets use test cases=> p=2->2 is a factor of 78 and 2 is a prime number. p=78->78 is a factor of 78 and clearly 78 isn't a prime number . Hence not sufficient. Hence A. Source->GMAT-prep

72)If the number 13 completely divides \(x\), and \(x = a^2 * b\), where a and b are distinct prime numbers, which of these numbers must be divisible by 169? A)\(a^2\) B)\(b^2\) C)\(a*b\) D)\(a^2*b^2\) E)\(a^3*b\)

A. There are only two numbers for which the number of factors = number itself. They are ->{1,2} As p>1 => p must be 2. So p=2 Clearly p is a prime number. Hence Sufficient. Statement 2->Using test cases p=2-> 2 is a factor of 26 and 2 is a prime number. p=26-> 26 is a factor of 26 and 26 is not a prime number. Clearly not sufficient. Hence A.

74)Data Sufficiency->Is the number of distinct prime factors of the positive integer N more than 4? (1) N is a multiple of 42. (2) N is a multiple of 98

E. Method 1-> 180 is a relatively smaller number=> Factors can be seen as follows -> 1*180 2*90 3*60 4*45 5*36 6*30 9*20 10*18 12*15 15*12(The repetition has begun. Hence checking the factors we get => There are 6 factors of the form 4k+2 for k≥0 Method 2-> 180=2^2^3^2*5 Factors of the form -> 4k+2 => 2(2k+1)=> 2*odd number. 2*1 2*3 2*3^2 2*5 2*3*5 2*3^2*5 Hence six values. E.

83)Data Sufficiency->If A and B are positive integers, does B have more prime factors than A? (1) B is the square root of A (2) A ÷ B is an integer

D. X and X^n have the exact same primes. So Statement 1 is sufficient. As for statement 2-> If A/B is an integer => Every prime factor of B must be a prime factor of A. Hence B can never have more prime factors than A. Hence Sufficient. Hence D.

84)If N = 10! then what is the minimum number which should be multiplied with N to make it a perfect square? A)7 B)14 C)21 D)28 E)36

A. 10!=1*2*3*4*5*6*7*8*9*10=> 2^8*3^4*5^2*7 In a perfect square the Power of every prime must be even. Thus =>Smallest number to be multiplied will be 7. Hence A.

85)If N = \(2^2*3^3*5^5\), find the total number of odd and even factors of N. A)24,96 B)15,30 C)24,48 D)15,96 E)15,48

C. For even factors => Power of 2 must be at-least 1. Hence possible cases => 2*4*6 =>48 odd factors => Power of 2 must be zero. Possible cases =>1*4*6=>24

Hence 48,24 is the correct option.

86)Which of the following numbers is a prime number? A)2343 B)3457 C)4689 D)7731 E)9861

B. Every number except the second one is dividable by 3 so they can never be prime. Hence B must be a prime number. Alternatively prime numbers can be written as either 6n+1 OR 6n-1 Only B can be written as 6n+1 One of the other options can be written as 6n+1 or 6n-1. Hence B is a prime number.

87)The product of all the prime numbers less than 20 is closest to which of the following powers of 10 ? (A) \(10^9\) (B) \(10^8\) (C) \(10^7\) (D) \(10^6\) (E) \(10^5\)

Product=>2*3*5*7*11*13*17*19=> 10*20*150*300=>9000000=> Close to 1 crore or 10^7 then it is close to 10 lakhs or 10^6 Hence C. Source->Official Guide.

88)If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have? (A) four (B) five (C) six (D) seven (E) eight

90)Data Sufficiency->What is the value of the integer p ? (1) Each of the integers 2, 3, and 5 is a factor of p. (2) Each of the integers 2, 5, and 7 is a factor of p.

C. Statement 1->3 and 5 must be the prime factors of K. Statement 2-> 2 and 5 must be the prime factors of K. combing them -> 2,3 and 5 must be the prime factors of K. Hence C. Source->Official Guide.

92)If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18

D. x->(-1.3,4) So possible integer values for x =>-1,0,1,2,3 Thus,five values for x are possible. Hence D.

96)For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301? A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150

97)If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ? (A) S/2 (B) S (C) 2S (D) 2S + 2 (E) 4S

C. The easiest way here to pick options. Let n=1 Sum=1 Sum of evens = 2 Hence for S=1=> Sum of evens = 2 Only Option C fits that case. NOTE-> Sometimes while using test cases more than one value may satisfy.In that case we must use some additional test cases to arrive at the correct answer. Source->2011 Official Guide.

98)How many prime numbers between 1 and 100 are factors of 7,150 ? (A) One (B) Two (C) Three (D) Four (E) Five

D. The question is asking us in a fancy way to get the prime factors of 7150 between 1 and 100. Breaking down 7150=> 2*5^2*11*13 Hence It has 4 prime factors. Hence D. Source-> Official Guide.

99)If n is a prime number between 0 and 100, how many positive divisors does n^3 have? A)1 B)2 C)3 D)4 E)5

102)If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n? A)10 B)11 C)12 D)13 E)14

103)n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is A)8 B)12 C)6 D)18 E)24

C. NOTE->If you choose A=> Have a relook at the Question."k is an even integer" is specified in the Question-stem.

106)If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ? (A) Two (B) Three (C) Four (D) Six (E) Eight

C. Easiest way is to put p=3 ->n=12 ->2,4,6,12 => Four even factors. Alternatively -> \(n=2^2*p\) where p is an odd prime => for even factors => power of 2 must be at-least 1. Possible cases => 2*2=> 4 Hence C.

107)If N = \(2^2 * 3^3 * 5^5\),How many factors of N are divisible by 5 but not divisible by 3. A)10 B)12 C)14 D)15 E)20

D. For factors to be divisible by 5 but not divisible by 3=> Power of 3 must be zero and power of 5 must be greater than or equal to 1. Possible cases => 3*1*5->15 Hence D.

108)What is the greatest prime factor of 12!11! + 11!10!? (A) 7 (B) 11 (C) 13 (D) 17 (E) 19

109)Data Sufficiency->How many distinct factors does positive integer k have? (1) k has more distinct factors than the integer 9 but fewer distinct factors than the integer 81. (2) k is the product of two distinct prime numbers.

D. Statement 1-> 9 has 3 factors and 81 has 5 factors => k must have 4 factors => Sufficient. Statement 2-> k=p*q for prime p and q => Number of factors => 2*2=>4=> Sufficient. Hence D.

110) What is the greatest prime factor of (2^29) - (2^26)? A)2 B)4 C)7 D)8 E)3

D. Statement 1-> Notice the boundary condition for n is given. Since n is an integer => n^2 must be a perfect square. n^2=> 49,64 or 81 So n can be 7 or 2^3 or 3^2 In each case -> n will have just one prime factor. Hence Sufficient.

Statement 2-> 8n^12 has twelve factors. 2^3*n^2 has 12 factors => This is only possible if n is a prime number. Thus it will have only one prime factor. Hence D. Source-> GMATClub-Tests

112)What is the greatest prime factor of 9919? A) 7 B) 13 C) 17 D) 97 E) 109

B. Two properties are been tested here => One-> \(X\) and \(X^n\) always have the exact same prime factors. Two->When we multiply any positive integer x with another positive integer n -> The number of prime factors of nx may be greater than or equal to x.It depends on the value of n. So \(x\) and \(x^2\) will have the exact same primes. Two cases are possible here => Case 1=> \(x\) has one prime factor which is not equal to 5. So 5\(x^2\) will have 2 primes. Case 2-> \(x\) has two prime factors,one out of which is 5 so that 5x^2 has two prime factors.

Hence almost \(x\) can have 2 prime factors. Hence B. Source->GMAT-prep

114)What is the greatest prime factor of 12!11! + 11!10!? (A) 7 (B) 11 (C) 13 (D) 17 (E) 19

E. 12!11! + 11!10!=> 11!10![12*11+1]=>11!10!*133=>11!10!*7*19 Clearly the greatest prime factor will be 19. Hence E.

115)Data Sufficiency->How many different prime numbers are factors of positive integer n? (1) 4 different prime numbers are factors of \(2n\). (2) 4 different prime numbers are factors of \(n^2\).

B. Properties in action-> \(X\) and \(X^n\) always have the exact same prime factors. When we multiply any positive integer x with another positive integer n -> The number of prime factors of nx may be greater than or equal to x.It depends on the value of n. Statement 1=> From this statement we can conclude that n can have 4 prime factors or 3 prime factors. Hence not sufficient. Statement 2=> From this statement it is evident that n must have exactly 4 prime factors. Hence Sufficient. Hence B. Source-> GMAT-Prep

116)What is the greatest prime factor of 3^6 - 1 ? A. 2 B. 3 C. 7 D. 13 E. 17

3^6-1=> 27^2-1=> (27+1)*(27-1)=> 2^3*7*13 Clearly 13 is the greatest prime factor. Hence D.

117)Data Sufficiency->If m and n are different positive integers, then how many prime numbers are in set {m, n, m + n}? (1) mn is prime. (2) m + n is even.

C. In this Question,we are told that m and n are different positive integers and we are asked as to how many numbers out of {m, n, m + n} are prime. Statement 1=> m*n is a prime number This can happen only when one out of m or n must be 1 and the other must be a prime number. Case 1 => When an even prime i.e 2 is involved -> {1,2,3} => two numbers in this set will be prime. Case 2=> When an odd prime is involved -> {1,odd-prime,even number >2}=> Only one prime number. Hence not sufficient. Statement 2->m+n is even e.g=> {3,5,8} => two prime or {2,4,6}=> None are prime => Not sufficient. Combining the two statements => m+n is even and one out of the is one => The other must be odd prime. Thus in the set => {1,Odd-prime,Even number>1}=> Only one number will be prime. Hence C.

118)What is the greatest prime factor of 1+2+3+。。。+40? A. 17 B. 29 C. 31 D. 37 E. 41

E. Sum => 1+2+3....40=> 20*41 => 2^2*5*41 41 is clearly the greatest prime factor. Hence E.

119)Data Sufficiency->How many positive factors does positive integer \(N\) have? 1) \(N^2\) has three positive factors. 2) \(2N\) has four positive factors.

A. We are asked about the number of factors of a positive integer N.

Statement 1-> \(N^2\) has 3 factors The only way \(N^2\) can have 3 factors is if N is a prime number. So \(N\) must have 2 factors. Hence Sufficient. Statement 2-> \(2N\) has 4 Factors Two cases are possible. Case 1=> \(N\) is a prime number.=> \(N\) will have 2 factors Case 1-> \(N=2^2\)=> \(N\) will have 3 factors.

Hence Not sufficient.

Hence A.

120)If \(n=3*4*p\) where p is a prime number greater than 3,how many different positive non-prime divisors does n have, excluding 1 and n? (A) Six (B) Seven (C) Eight (D) Nine (E) Ten

B. \(n=3*2^2*p\) n will therefore have 12 factors including1,n and its prime factors 2,3,p So excluding them all n will have -> 12-5=7 factors. Hence B

121)Data Sufficiency->How many divisors does the positive integer \(N\) have? (1)\(27N^3\) has 16 factors. (2)\(90<N^3<200\)

D. In statement 1 we can collude that N is a prime number. In statement 2 => N must be 5. Hence D.

122)Data Sufficiency-> How many prime factors does positive integer n have? (1) n/5 has only a prime factor. (2) 3*n^2 has two different prime factors.

126)Data Sufficiency->If \(p\) is a positive integer, is \(p\) a prime number? (1) \(p\) and \(p+1\) have the same number of factors. (2) \(p-1\) is a factor of \(p\).

B. We need to check if p is prime or not. Statement 1->B. Lets use some test cases => p=2 p+1=3 Both have 2 factors as both are prime. Hence p is prime But wait.Lets find some other test case for which p is non prime. p=21 p+1=22 Both have 4 factors And p is clearly non prime. Another case be 14 and 15 Hence not sufficient. Actually there exist ∞ such cases. Statement 2-> For every positive integer >1 p and p-1 are always co-prime. The only value of p possible is 2. Hence p=2,which is a prime number. Hene B

127)Data Sufficiency->If p is a positive integer, is integer k prime? (1) 3p + 3 = k (2) 7! + 3 = k

C. From statement 1->Using test cases=> k =4 or k=3 Hence not sufficient. From statement 2=> k=3 or k=15 Hence not sufficient Combing them => k+7 will be an even number >2, so it cannot be prime. Hence k must be prime so that one out of 1,k,k+7 is prime. Hence C.

129)Data Sufficiency->What is the value of the integer p? (1) p is a prime number. (2) 88 ≤ p ≤ 95

E. Note-> For combination statement=> Two examples for n=31^2=> n has less than 4 factors or n=900=> n has more than 4 factors

131)If x is an integer and x^2 is even, which of the following must be true? I. x is odd. II. x is even. III. x^3 is odd. (A) I only (B) II only (C) III only (D) I and II only (E) II and III only

B. Property in action=> x and x^n always have the same even/odd nature if n≠0

132)Data Sufficiency->If K is a positive 3-digit number, is K prime? (1) The last digit of K is not even (2) K is the smallest number possible where the hundreds digit is the sum of the tens and units digit. Tens and units digits are equal.

B. Statement 1 is off course nor sufficient. We can use test cases => 211 is prime and 145 is not prime. Notice that from statement 2 => K must be 211. And 211 is a prime number as it is not divisible by any primes less than or equal to the square root of 211. Hence B.

133)Data Sufficiency->If x and y are positive integers, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x. (2) x is an odd number

A. It is all about patterns. Here is what i did in this Question=>

We need to see if y/3 is an integer or not. We are given that a and y are positive integers. Statement 1=> \(y = 2x^3 + 9x^2 - 5x\) Taking out x as a factor we get -> \(x[2x^2+9x-5]=> x(2x-1)(x+5)\) Now putting in the values of x=> y is always a multiple of 3. Hence sufficient.

Statement 2=> No clue of y=> Not sufficient.

Hence A.

134)If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

A) If \(x\) is even then \(y\) is odd. B) If \(x\) is odd then \(y\) is a multiple of 4. C) If \(x+y\) is odd then \(y/x\) is not an integer. D) If \(x+y\) is even then \(x/y\) is an integer. E) \(x^y\) is even.

136)Data Sufficiency->If x is a prime number, what is the value of x? (1) 2x + 2 is the cube of a positive integer. (2) The average of any x consecutive integers is an integer.

E. We are told that x is a prime number and asked about its value.

Statement 1-> 2x+2=t^3 for any positive integer t. Lets use some test cases => 2x+2=1 x=1/2=> Not allowed 2x+2=8 x=3=> Allowed 2x+2=27 x=25/2=> Not allowed 2x+2=64 x=31=>Allowed

Hence 3 and 31 are both acceptable values => Not sufficient.

Statement 2-> Remember for set of consecutive integers is an AP series. Mean can be of the form p if x is odd (p is an integer ) Mean can be of the form p.5 if x is even (p is an integer) RULE->Sum of n consecutive integers is always divisible by n for n being odd and never divisible by n for n being even. Hence this statement tells us that x is odd. But their are ∞ odd primes => Not sufficient. Combing them x=3 and x=31 are both acceptable values. Hence not sufficient .

Hence E.

137)The sum of prime numbers that are greater than 60 but less than 70 is (A) 67 (B) 128 (C) 191 (D) 197 (E) 260

B. Only 2 prime are present in the boundary -> (60,70) => 61 and 67. Sum =128. Hence B. Source->Official Guide

138)Data Sufficiency->If x, y, and z are positive integers, what is the greatest prime factor of the product xyz? (1) The greatest common factor of x, y, and z is 7. (2) The lowest common multiple of x, y, and z is 84.

B. Given data -> x,y,z are positive integers. We are asked about the Greatest prime factor of x*y*z

Statement 1-> GCD(x,y,z)=7 Lets use test cases-> Case 1=> 7 7 7 GCD=> 7 And Greatest prime =7 Case 2=> 7 7 7*13 GCD=>7 Greatest prime factor is 7*13

Actually this statement tells us that the greatest prime factor must be greater than or equal to 7. Hence not sufficient.

Statement 2-> LCM(x,y,z)=84=2^2*3*7 Now if x,y,z had any prime factor other than 2,3,7 => It must have been present in the LCM. In LCM => We pick the greatest power of each Prime factor. Hence the greatest prime in the Product must be 7. Hence sufficient. Hence B.

139)If n=6p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n? A)2 B)3 C)4 D)6 E)cannot be determined.

E. Two cases exist here=> Case 1-> p=3=> Even divisors ->3 Case 2-> p=prime >3 => even divisors ->4 Hence E.

140)Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

A. a^3*b^4*c^5 B. a^5*b^4*c^3 C. a^2*b^3*c^5 D. a^7*b^6*c^5 E. a^27*b^26*c^25

E. Perfect square => Powers of all primes=> Multiple of 2. Perfect cube=>Powers of all the primes => Multiple of 3. Perfect fifth power => Powers of all the primes => Multiple of 5

Hence powers of all the primes must be a multiple of all->2,3,5 => LCM=30 Smallest multiplication => \(a^{27}*b^{26}*c^{25}\) Hence E.

141)Data Sufficiency->Is the integer b divisible by 6 ? (1) 8b is divisible by 3. (2) 9b is divisible by 12.

C. We need to see if B is divisible by 6 or not. Statement 1=> From this statement we can conclude that B must be divisible by 3. E.g-> 3,6 Not sufficient. Statement 2=> Form this statement , we can conclude that B must be divisible by 4 E.g->4,4*6 Not sufficient. Combining the two statement => B must be divisible by 12. Hence it must be divisible by 6 too.

Hence C.

142)Data Sufficiency->If k is a positive integer, is k a prime number? (1) No integers between 2 and \(\sqrt{k}\), inclusive divides k evenly. (2) No integers between 2 and k/2 inclusive divides k evenly, and k is greater than 5.

D. In order to check whether a given number is prime or not =>We must check its divisibility with all the prime numbers less than or equal to the square root of that number. So clearly A is sufficient.

Statement 2 is a bit a trickier. Lets compare the values of √k and k/2 for integers k>5 10=> √10=3.something 10/2=5 100=> √100=10 100/2=50 10000=> √10000=100 10000/2=5000 Clearly as the number is increasing the gap between √k and k/2 is also increasing. Hence k must not have any prime factors between 1 and √k too.[/highlight]

Thus k is a prime number. Hence Sufficient.

Hence D.

143)The "prime sum" of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example , the prime sum of 12 is 7, since 12 = 2 x 2 x 3 and 2 +2 + 3 = 7. For which of the following integers is the prime sum greater than 35 ? (A) 440 (b) 512 (C) 620 (D) 700 (E) 750

E. We are told that A and B are positive integers and we are asked if if B/A is an integer or not. Lets use some test cases here.

Statement 1-> 2B/A is an integer. Let A=1 and B=1 => 2B/A is an integer and B/A =1 which is an integer. Let A=2 and B=1 => 2B/A is an integer and B/A=1/2 which is not an integer. Hence not sufficient.

Statement 2-> B^2/A is an integer. Let A=1 and B=1 => B^2/A is an integer and B/A=1 which is an integer. Let A=4 and B=2 => B^2/A is an integer and B/A=1/2 which is not an integer. Hence not sufficient.

Combing the two statements -> Let A=1 and B=1 => B/A is an integer. Let A=4 and B=2 => B/A is not an integer. Hence not sufficient. Hence E.

146)Data Sufficiency->If n is a positive integer is n-1 divisible by 3 ? (1) n^2+n is not divisible by 6. (2) 3n=3k+3 where k is a positive multiple of 3

147)Data Sufficiency->If n is a positive integer, is n – 1 divisible by 3? (1) n^2 + n is not divisible by 6. (2) 3n=k+3 where k is a positive multiple of 3.

148)Suppose x is the product of all the primes less than or equal to 59. How many primes appear in the set {x + 2, x + 3, x + 4, …, x + 59}? A)0 B)17 C)18 D)23 E)24

149)The smallest prime factor of 899 is \(x\). Which of the following is true of \(x\)? A. \(1 \lt x \le 7\) B. \(7 \lt x \le 14\) C. \(14 \lt x \le 21\) D. \(21 \lt x \le 28\) E. \(28 \lt x \le 35\)

E. We need to see if the number of factors of n is greater than or equal to 4 or not.

A few key links =>

1 is the only number that has one factor.

A prime number has exactly two factors.

For any number to have 3 factors it must be of the form => Prime^2 i.e square of a prime number.

Lets get on with the statements=>

Statement 1-> n is not prime. Lets play around with some test cases. n=1 => One factor => n has less than 4 factors n=200=> 2^3*5^2 => 12 factors => n has more than 4 factors. Hence insufficient.

Statement 2-> n lies in the range [150,200] Again lets use some test cases here. n=151=> Prime number as it is not divisible by any prime number less than equal to the square root of 151. So n has 2 factors which is less than 4. n=200=> 12 factors => more than 4 factors. Not sufficient.

Combing the two statements => n is not prime. ans n≠1 So n can never has one or two factors.

But what about three? Can n be square of a prime ? YES for n=13^2=169=> n has 3 factors which is less than 4 For all values of n => n will have more than 3 factors. Hence not sufficient. Hence E.

151)Data Sufficiency->If p is a positive integer, what is the value of p? (1) p/4 is a prime number. (2) p is divisible by 3

153)An integer n that is greater than 1 is said to be "prime-saturated" if it has no prime factor greater than or equal to √n. Which of the following integers is prime saturated? A) 6 B) 35 C) 46 D) 66 E) 75

155)If x is a positive integer greater than 1, what is the sum of the multiples of x from x to x^2, inclusive? (A) x(x + 1)(x-1) (B) x^2(x + 1)/2 (C) x^2(x-1) (D) (x^3 + 2x)/2 (E) x(x-1)^2

B. Plugging in x=5 => Sum =75 Only option that matches is B. Hence B. Note=> While plugging in numbers we must check for each option.There are cases when you might get more than one answer. In such cases we must alter our test cases to choose between the remaining options.

156)Data Sufficiency->Does p^2 = q if p is a prime number? (1) q^2 – p^2 =0 (2) p^2 = 49

157)A number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n. What is the greatest two digit prime saturated integer ? A. 99 B. 98 C. 97 D. 96 E. 95

E. Key Properties applicable in this question => X and X^n always have the exact same prime factors.

When we multiple a positive integer x with some other positive integer n => The number of prime factors of nx may be greater than or equal to the number of prime factors of x. We need the number of prime factors of x. Statement 1-> 5x^2 has primes x can have 2 primes (if 5 is one of its primes) x can have 1 prime (if 5 is not one of its primes) Hence not sufficient. Statement 2-> x>1 Clearly not sufficient. Combing the two statements => Still not sufficient.

E.g => x=3 or x=5*3 Satisfy both the equations.

Hence E.

159)How many different prime factors does positive integer x have? (1) \(1 < x < 6\) (2) \(5x^2\) has four factors.

D. We need the number of prime factors of the positive integer x. Statement 1=> x=>(1,6)=> x=2,3,4,5=> In each case => x will have just one prime factor=> Sufficient. Statement 2=> The only value of x possible is 5. Hence x will have only one factor. Sufficient

Hence D.

160)Data Sufficiency->If N is a positive integer, does N have exactly three factors? (1) The integer N^2 has exactly five factors (2) Only one factor of N is a prime number

Firstly a few key kicks => 1 is the only number that has one factor. A prime has exactly 2 factors. For a number to have 3 factors =>It must be of the form => Prime^2 i.e square of a prime number.

Lets jump into Statements. Statement 1-> n^2 has 5 factors. This can only happen if n=Prime^2 Hence sufficient.

Statement 2-> This statement is basically saying that n has only prime factor. But it does not tell us anything about the exponent of that prime. E.g=> n=2^2=> 3 factors n=2^100=>101 factors Hence not sufficient. Hence A.

161)Data Sufficiency->What is the value of x? (1) x is the square of an integer. (2) 577 < x < 675

162)Data Sufficiency->If x is a prime number, what is the value of x. 1)There are 4 prime numbers between 11 and x 2 )There is no y such that 1<y<x and y is the divisor of x.

164)For how many positive integers is the number of positive divisors equal to the number itself ? A) none B) one C) two D)three E)cannot be determined

166)Data Sufficiency->If x is a perfect square greater than 1, what is the value of x? (1) x has exactly 3 distinct factors. (2) x has exactly one positive odd factor

We are given that x is a perfect square and are asked its value. Statement 1-> x has 3 factors. Some quick facts on factors -> 1 is the only number that has 1 factor. A prime as 2 factors. A number of the form Prime^2 has 3 factors. Hence x must be of the form Prime^2 E.g=> 2^2 or 3^2 etc. Hence not sufficient.

Statement->2 As 1 is the factor of every number => This statement tells us that x must be even. E.g => 2^2 or 2^4 or 2^6 Basically it cannot have any prime factor other then one. So it must be of the form 2^even number . Hence not sufficient.

Combing the two statements=> x must be 2^2 to have 4 factors. Hence sufficient. Hence C.

167)Data Sufficiency->How many factors does y have? (1) y is the cube of an integer. (2) y is the product of 2 distinct positive digits.

Statement 1-> y is a perfect cube. y can be negative or zero or positive. It would have been nice if the Question Stem had read-> y is a polite integer. Anyways its clearly not possible to get the factors. Not sufficient.

Statement 1-> From this statement we can infer that y must be positive. y=2*3=> 6(4 factors) y=1*5=>5(two factors) y=3*9=>27=> 4 factors. Hence not sufficient.

Combing the two statements=> The only values that y can take are 8 and 27 and they both have 4 factors.

Hence C.

168)Data Sufficiency->If x, y and k are integers, is xy divisible by 3? (1) y = 2^(16) - 1 (2) The sum of the digits of x equals 6^k

E. Note-> x=2 and x=7 satisfy both the statements.

170)Data Sufficiency->How many prime factors does positive integer n have? (1) n/7 has only one prime factor. (2) 3*n^2 has two different prime factors.

E. Note-> Values such as x=7^2 and 7*3 will satisfy both the statements. Hence E.

171)If x = 13y, where y is a prime number greater than 2, how many different positive even divisors does x have, including x? A. 0 B. 1 C. 2 D. 3 E. It cannot be determined from the information given

172)Data Sufficiency ->If b is a positive integer, then b has how many distinct positive factors? (1) all of b's factors are also factors of 62. (2) b is the product of one even prime number and one odd prime number.

Statement 1-> This is just another way of saying that 62/b will be an integer i.e b is a factor of 62. b=2 b=1 b=62 Etc . Hence not sufficient.

Statement 2=> b=Product of two prime numbers that are different. Hence b will be of the from prime1 *prime2 Hence number of factors will be 4.

Hence sufficient. Hence B.

173)Data Sufficiency->How many different factors does the integer n have? (1) n = (a^4)(b^3) where a and b are different positive prime numbers. (2) The only positive prime numbers that are factors of n are 5 and 7.

A. We need to get the number of factors of positive integer n.

Statement 1=> As a and b are "different" prime numbers => Number of factors of a must be 5*4=20 Hence sufficient . Statement 2=> There exist ∞ numbers with the same set of prime numbers. E.g 5*7=> Four factors. 5^2*7^2=> Nine factors. Etc. Hence not sufficient.

Hence A.

174)Data Sufficiency->If x and y are positive integers and x + y = 3^x, is y divisible by 6? (1) x is odd. (2) x is a multiple of 3.

Given info->x and y are positive integers. x+y=3^x In order for y to be divisible by 6 => It must be divisible by both 2 and 3. Statement 1-> Test Case 1-> x is odd x=1 y=y x+y=3^1=> Yes Clearly y is not divisible by 6. Test Case 2-> x =3 y=24 x+y=3^3=>Yes Clearly y is divisible by 6 Hence not sufficient. Statement 1-> x is a multiple of 3. So y must be a multiple of 3 too. Now=> Test case 1-> x=3 y=24 x+y=3^3=>Yes Clearly y is divisible by 6. Test case 2-> x=6 y=3^6-6=>Odd number -Even number =>Odd number => Never divisible by 6. Hence not sufficient. Combing the two statements=> From Statement 1=> y=> even From Statement 2=> y must be a multiple of 3. Hence y must be a multiple of 6. Hence Sufficient. Hence C.

175)Data Sufficiency ->If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.

D. As n=x^3-x=> (x-1)*x*(x+1)=> If x is odd => both x-1 and x+1 will be consecutive even integers.So one of them would be multiple of 4 and other would be a multiple of 2 for sure.Hence n would be a multiple of 8.

Statement 1-> 3x is odd => x is odd => Sufficient. Statement 2-> x=4y+1 => Even+odd=odd Hence Sufficient.

Hence D.

176)If a positive odd integer N has p positive factors, how many positive factors will 2N have ? A) p B) 2p C) P+1 D) 2p+1 E) Cannot be determined

B. We are told that n is an integer. Statement 1-> n/3=integer. E.g=> n=3 or n=6 etc. Hence not sufficient. Statement 2-> If n is odd => The number of factors of 2n will always be 2 times the number of factors of n. This is because any odd number will just have odd divisors.It cannot have any even divisor as 2 is not its prime factor. After we multiple 2 to n => The numb of factors will get doubled as for every odd factor there would be an even factor.

Hence n is odd Thus-> This statement is sufficient. Hence B.

178)If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following?

I. 5 II. 2x III. x+1

A. I only B. II only C. III only D. I and II only E. II and III only

D. All prime numbers greater than 2 are odd. Hence p must be of the form => 2k+1 p^2=> 4k^2+4k+1=> 4k(k+1)+1= 8k'+1 Hence it will always leave a remainder 1 with 8.

Another important takeaway -> For any odd number k => k^2 will always leave a remainder 1 with 8.

180)Data Sufficiency->Is positive integer N divisible by 3?

(1) N^2/36 is an integer (2) 144/N^2 is an integer

A. A very useful property -> X and X^n always have the exact same prime factors. For N to be divisible by 3 => 3 must be the prime factors of N.

Statement 1=> N^2/36 is an integer. N^2 has both 2 and 3 as it prime. Thus N must have 3 as its prime too. Hence Sufficient. Statement 2=> Lets use test Cases here. N=1 => 144/1^2 =Integer => 1 is not divisible by 3. N=3 => 144/N^2 =integer => 3 is divisible by 3. Hence not sufficient.

Hence A.

181)Data Sufficiency->If x is an integer, is x^3 divisible by 7? (1) 3*x^12 is divisible by 7 (2) 3*x^4 is divisible by 7

A. Firstly n^3-n= (n-1)n(n+1) It is a product of 3 consecutive integers. Regardless of what the value of n is => It will always be a multiple of 6. Reason -> Product of n consecutive integers is divisible by n!

Now lets see two cases ->

Case 1-> n is odd If n is odd => n-1 will be even and n+1 will be even too. Further more one out of them will be multiple of 4. Thus (n-1)n(n+1) will be multiple of 8. As it is already a multiple of 6 => we can conclude that (n-1)n(n+1) will be a multiple of 24.

Case 2-> n is even In that scenario it is not possible for us to conclude anything except (n-1)n(n+1) will be multiple of 6.

Hence we just need to see if n is odd or not.

Statement 1-> n=2k+1 for some integer k. Hence n=even+odd=odd Hence n must be a multiple of 24. Hence sufficient.

Statement 2-> Lest use some test cases=> n=2 => (n-1)n(n+1) = 1*2*3 => Not a multiple of 4. n=3 => (n-1)n(n+1)=2*3*4 =>Multiple of 24.

Hence not sufficient.

Hence A. Source->Official Guide.

186)For which of the following values of n is (100+n)/n NOT an integer? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

A. We need to see if a*b*c is divisible by 24 or not.

Statement 1-> a,b,c are consecutive evens =>Let the consecutive evens be => 2n 2n+2 2n+4

Taking the product => 2n(2n+2)(2n+4)=> 8n(n+1)(n+2)

Product of t consecutive integers is always divisible by t! Hence 8n(n+1)(n+2)=8*6k = 48k for some integer k. Clearly it will be divisible by 24. Hence sufficient.

Statement 2-> Here we have no clue whether a,b,c are integers or non integers. Hence not sufficient. Hence A.

188)Data Sufficiency->If x is the product of integers a, b, c, and d, is x divisible by 128? (1) a = 24 (2) a, b, c, and d are consecutive even integers

B. We are told that a,b,c,d are all integers. We need to see if x=a*b*c*d is divisible by 128 or not.

Statement 1-> Lets use test cases. Case 1=> a=24 b=1 c=1 d=1 x=24 => Clearly it is not divisible by 128. Case 2=> a=24 b=1 c=1 d=128 x=24*128 => divisible by 128. Hence not sufficient.

A. From statement 1-> N must be a prime number.Hence it will always have 2 factors.Hence sufficient. From statement 2 -> Two cases are possible => N=5^2 and N=Any prime number other than 5. Hence N can have 3 or two factors. Hence not sufficient.

Hence A.

190)Which of the following statements must be true? A)If n is even then n^3-n is divisible by 24. B)The sum of first 1000 prime numbers is even. C)The product of first 32 primes is odd. D)2 and 3 are the only consecutive prime numbers. E)If n is odd then n^3-n is always divisible by 24.

E. Mote-> D is not true as there is a BIG difference between consecutive prime numbers and consecutive numbers that are prime. So 5,7 or 11,13 or 17,19 etc are all consecutive prime numbers.

2,3 are the only consecutive integers that are also prime numbers.

191)Which of the following is NOT a factor of 10! ? A. 1440 B. 625 C. 160 D. 80 E. 50

C. We need to get the number of prime factors of N.

Statement 1-> N is a factor of 7200 7200=2^5*3^2*5 Hence N can have --> Zero prime factor if N=1 One prime factor Two prime factors Or Three Prime factors. Hence not sufficient.

Statement 2-> 180 is a factor of N. Hence N=2^2*3^2*5*k for some integer K. Hene N must have 2,3,5 as its prime factors. Other than that N can have any Prime factors ≥3 Hence not sufficient.

Combining the two statements => N must have exactly three prime factors> Hence sufficient.

Hence C.

194)Data Sufficiency->If F is the prime factorization of N!, how many factors in F have an exponent of 1? (1) 30 ≤ N ≤ 40 (2) 25 ≤ N ≤ 35

195)Data Sufficiency->Given that \(x\) is an integer and \(x\) is positive, is \(\frac{210}{x}\) also an integer?? (1) \(x\) is a prime number. (2) \(x\) < 8

Observe that 210=2*3*5*7 We need to see if 210/x is an integer or not. Statement 1-> x is prime E.g=> x=13 => No x=7 => Yes Hence not sufficient.

Statement 2-> x<8 E.g-> x=4 -> No x=7 => Yes Hence not sufficient. combining the two statements => x can be 2,3,5,7 Each one if them is a factor of 210. Hence 210/x will always be an integer. Hence sufficient. Hence C.

196)Which of the following is NOT a factor of the product of the first 50 positive multiples of 4 ? A. \(17^2\) B. \(11^4\) C. \(7^6\) D. \(47^{12}\) E. \(2^{124}\)

D. Firstly the product of first positive 50 multiples of 4=> 4*1*4*2*4*3*4*4*4*5.....4*50=> 4^50*50!

Secondly notice Option D is 47^2 and 47 is a prime number.

Number of 47's in 4^50*50!=> one

Hence 47^2 will never be its factor.

Hence D.

197)Data Sufficiency->Is the integer \(n\) odd? (1) a is an integer and \(n=a^7+a^5+a^3+a^2+2a+1\) (2) 2n is divisible by twice as many positive integers as n

D. Statement 1-> Here a can be even or odd => n will always be odd. Hence sufficient. Statement 2-> If n is odd => The number of factors of 2n will always be twice the number of factors of n. Hence n is odd => Sufficient.

Hence D.

198)If a positive integer n has 211 factors,then how many prime factors does n have? A)one B)two C)three D)Four E)cannot be determined

A. Notice here that 211 is a prime number as it is not divisile by any prime factor less than or equal to the square root of 211. Therefore the number n can only be of the form => Prime^210 Hence it will always have one prime factor.

If the number of factors of any integer is a prime number => The integer will always have a single prime factor. Hence A

199)If a positive integer N has p factors ; how many factors will 2N have ? A) p B) 2p C) P+1 D) 2p+1 E) Cannot be determined

Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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Day-3

Mock Test 3

Number of Questions->85

Topic Covered-> Evens/odds

Source-> This Quiz contains questions from varied sources along with quite a few self-made questions.

1)Which of the following statements must be true? A)Any integer can be either even or odd but not both. B)2n is always even. C)2n+1 is always odd. D)Zero is an even integer. E)The sum of even number of odd numbers is always is always even. F)The sum of odd number of odd numbers is always odd. G)If any factor of an integer is even,then the integer itself would be even.

4)If x and n are non negative integers,then which of the following statements must be true? A)If x is even then x^n will always be even. B)If x is odd then x^n will always be odd.

B. Notice that n can be zero too. Although x and x^n will always have the same even/odd nature,we must make sure than the exponent is non zero. Anything^zero=1 which is always odd.

5)Data Sufficiency->If \(A\) and \(B\) are integers,is \(A*B^4\)even? A)\((2A+B)^3\) is even. B)\(A+3B\) is odd.

6)If w,x,y,z are consecutive positive integers then which of the following statements will always be odd? I) wx+(y+2)^2*z II) w^x+z^y III) 2w^3-3x^2-5y+6z^2 A)I B)II C)III D)I and II E)II and III

D. Essentially this is testing our knowledge on Even/odd numbers. If sum of 2 primes is even => 2 must not be either of them. If sum of 2 primes is odd => 2 must be one of them.

Hence as 79 =2+77 =>79 can never be written as sum of 2 prime numbers.

Hence D.

8)Data Sufficiency->If \(x\) and \(y\) are integers, is \(3*x^4 + 4*y\) even? (1) \(x^3\) is even. (2) \(y^{2x} + 3\) is even.

B. Note-> Statement 1 is not sufficient as we gotta consider the case of x=0.

10)Data Sufficiency->If \(X\) is a positive integer, is \(X^2 + 1\) an odd number? (1) \(X\) is the smallest integer that is divisible by all integers from 11 to 15, inclusive. (2) \(3^X\) is an odd number.

11)Data Sufficiency-> If a, b, and k are positive integers, is the sum (a + b) an even number or an odd number? (1) a = ( k^3 + 3k^2 + 3k + 6) (2) b = (k^2 + 4a +5)

12)Data Sufficiency->If x, y, and z are positive integers, where x is an odd number and z = x^2 + y^2 + 4. Is y^2 divisible by 4? (1) Z = 8k -3 where k is a positive integer. (2) When (z-x+1) is divided by 2, it leaves a remainder.

13)If a, b, c, d, and e are integers and the expression \(\frac{a*b^2*c^2}{d^2*e}\) gives a positive even integer, which of the following options must be true? I. \(a*b*c\) is even II. \(\frac{a}{e}\) is positive III. \(\frac{a}{d^2}\) is positive.

15)If r and s are positive integers, and \(r^2\) + \(\frac{r}{s}\) is an odd integer, which of the following cannot be even? A)\(3*r+2*s\) B)\((r-1)*(s+2)\) C)\(r^{s+1}+s^r\) D)\(r^3+3\) E)\(s^4+4\)

18)If A is a positive integer, then which of the following statements is true? 1. A^2 + A -1 is always even. 2. (A^4+1)(A^4+2) + 3A is even only when A is even. 3. (A-1)(A+2)(A+4) is never odd.

20)Set S is given as S = {1,3,5,7,9,11,13,15,17}. In how many ways can three numbers be chosen from Set S such that the sum of those three numbers is 18? A)zero B)two C)three D)six E)nine

A. Notice that all the integers in the set are odd So whenever we pick any 3 numbers => Sum will be odd+odd+odd=> odd Hence sum will never be 18 which is even. Hence A.

21)Data Sufficiency ->If A and B are positive integers,is the Product A*B even? A)A is odd B)B^43 is even

22)Which of the following statements must be true? The product of first 100 prime numbers is even. The sum of first 100 prime numbers is odd. The sum of first five non-negative even numbers is divisible by both 4 and 5.

A)I only B)II only C)I and II only D)I and III only E)I, II and III

E. Properties in action => Sum of odd number of odd numbers -> Odd Sum of even number of odd numbers is even. Two is the only even prime.

23)If p,q,r,s are consecutive positive integers,then which of the following statements must be odd ? I)p*s + [(q+14)^2] *r II)p^r + s^q III)26*p^3 + 3*q^23 + 5*r^21 + 28*s^2

A)I only B)II only C)I and II only D)II and III only E)I, II and III

27)Data Sufficiency->If \(x\) and \(y\) are positive integers and \(x\) is odd, is \(x*y\) even? (1) \(x^3*y = 6*a^3 + 23\) where a is a positive integer. (2) \(x^2+y = 3*k + 7\) where \(k\) is a positive integer.

30)If n is an integer, then which of the following statements is/are FALSE? I)\(n^3 – n\) is always even. II)\(8n^3 +12n^2 +6n +1\) is always even. III)\(√ (4n^2 – 4*n +1)\) is always odd.

A)I only B)II only C)I and II only D)II and III only E)I, II and III

B. Notice that for statement 3-> 4n^2-4n+1 is (2n-1)^2

31)Data Sufficiency ->If t is a positive integer, is t^3 + 1 an odd number? (1) t is the smallest integer that is divisible by all integers from 21 to 25, inclusive. (2) 5^t is an odd number.

A. Also notice that in statement 2,its not a necessity that y will be odd.If x=0 then y can be even or odd.This fact is useless in this questions as we don't want the even/odd nature of y,rather we are only interested in the even/odd nature of x.

33)Data Sufficiency ->If p, q, and r are positive integers, where p is an odd number and r = p^2 + q^3 + 4. Is q^3 divisible by 8? (1) r = 18k -5 where k is a positive integer (2) When (r-p+13) is divided by 2, it leaves a remainder.

D. Notice that this question is just asking us for the even/odd nature of q. If q is even => q^3 must be divisible by 8.

34)Data Sufficiency ->If a, b, and t are positive integers, is the sum (a + b) an even number or an odd number? (1) a = ( t^13 + t^12 + 3t+ 6) (2) b = (t^2 + 4a +5)

39)If a and b are positive integers, and a^3 + a^2/b is an odd integer, which of the following cannot be even? A)31a + 12b B(a-1)(s+10) C)ab+21b D)a^3 +7 E)b^4 +4

D. Notice that if a^3 + a^2/b is odd -> a must be even and b must be even too.

40)Set S is given as S = {1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37}. In how many ways can three numbers be chosen from Set S such that the sum of those three numbers is 68? A)zero B)1 C)19 D)38 E)19!

A. Notice that they are all odd. Hence, sum of any three integers picked from this set will be odd + odd +odd -> odd Hence sum can never be 68. Hence A.

41)Data Sufficiency->If a+b is even, is b an integer? 1) a-b is even. 2) a+3b is even.

42)Which of the following statements must be true? I)The product of first 249 prime numbers is even. II)The sum of first 249 prime numbers is even. III)Sum of any 13 prime numbers greater than 100 is always odd.

A)I only B)II only C)I and II only D)I and III only E)I, II and III

E. Two properties are in action here. ONE-> Sum of even number of odd numbers is even and the Sum of odd number of odd numbers is odd. TWO-> All primes numbers greater than 2 are odd.

43)State True/False->The sum of first 5 non negative of 5 is divisible by both 25 and 50. A)True B)False

A. The only value of a and b that will satisfy statement 1 is 7. Hence a=b=7

57)Data Sufficiency->If x, y and z are integers and xy + z is an even integer, is x an even integer? (1) xy + xz is an even integer. (2) y + xz is an odd integer.

60)If a is an odd integer and b is an even integer, which of the following CANNOT be true? A. a + b is an odd integer. B. a is a factor of b. C. b is a factor of a. D. a^b is an odd integer. E. b^a is an even integer.

C. An even number can never divide an odd number. Hence C.

61)If x is a positive odd integer and y is a negative even integer, which of the following must be true? A. x^3 + y is a positive odd integer B. x^2 + y^2 is a negative odd integer C. x^0 + y^11 is a negative odd integer D. x + y is a positive odd integer E. x + y is a negative odd integer

62)If x is a positive odd integer and y is a negative even integer, which of the following must be false? A. x^3 + y is a positive odd integer B. x^2 + y^2 is a negative odd integer C. x^0 + y^11 is a negative odd integer D. x + y is a positive odd integer E. x + y is a negative odd integer

67)If x and y are integers and x + y = 5, which of the following must be true? A) x and y are consecutive integers. B) If x < 0, then y > 0. C) If x > 0, then y < 0. D) Both x and y are even. E) Both x and y are less than 5.

D. We are told that x^2*y<0 Hence x≠0 and y≠0 So as x^2 is always greater than 0. Hence y<0. Also as x^2*y is odd => x and y must be both odd.

Option 1->xy^2=> odd*odd^2=> odd. True. Option 2-> xy will be negative for x being positive and xy will be positive for x being negative. Hence this statement is not always true. Option 3-> x+y=> odd+odd=> even. Hence this is True.

So the correct answer is D.

73)If x is an odd integer and y and z are even integers, which of the following CANNOT be an integer? A. y/z B. x/y C. z/x D. yx/z E. zx/y

82)If a is an odd integer, which of the following must be an even integer? A. a^4−a+1 B. (a^4−a)(a+1/a) C. a^4−a^3+a^2+2a D. (a^3+a^2+a)^2 E. None of the above.

B. Firstly -->Positive exponents does not affect the even/odd nature of any number Option A-> a^4−a+1=> odd-odd+odd=> even+odd=> odd=> Rejected. Option B->(a^4−a)(a+1/a)=> (a^3-1)*(a^2+a)=> (odd-odd)*(odd+odd) => even*even => even=> Acceptable. Option C. a^4−a^3+a^2+2a=> odd-odd+odd+even=>even+odd=>odd=> Rejected. Option D. (a^3+a^2+a)^2=>(odd+odd+odd)^2=>(even+odd)^2=>odd^2=> odd=>Rejected. Option E. None of the above.=> Rejected.

83)If a and b are odd integers, which of the following must be an even integer? A. a^2(b – 2) B. ab + 40 C. (a + 22)(b – 42) D. 31a + 51b E. a(a + 16)

85)If x is even integer, which of the following must be an odd integer? A. \(\frac{3x}{2}\) B. \(\frac{3x}{2} + 1\) C. \(3x^2\) D. \(\frac{3x^2}{2}\) E. \(\frac{3x^2}{2} + 1\)

2)If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value? A. 3x – 1 B. x^2 + 1 C. 3 – x D. x – 3 E. x^2 – x

A. We need the maximum absolute value. So actually we just need the maximum magnitude as the sign won't matter. Option 1->Maximum at x=-2 => Value=-7 =>Magnitude =7 Option 2->Maximum at x=-2 or 2=> Value=5 =>Magnitude =5 Option 3->Maximum at x=-2=>Value=5=>Magnitude =5 Option 4->Maximum at x=-2=>Value=-5=>Magnitude=5 Option 5->Maximum at x=-2=>Value=6=>Magnitude=6 Hence A.

3)For what values of x is the expression |\(x^2-3\)| minimum?

15)Data Sufficiency->If \(p\) and \(q\) are positive integers and \(X = 6^p + 7^{q+23}\), what is the units digit of \(X\)? (1) \(q = 2p – 11\) (2) \(q^2 – 10q + 9 = 0\)

16)Data Sufficiency->The number \(x\) is a positive odd integer. If the unit digit of \(x^3\) is subtracted from the unit digit of \(x^2\), it results in 0. What is the unit digit of the number \(x + 7\)? (1) The unit digit of the product of 105 and \(x\) is 5. (2) When x is divided by 5, it leaves no remainder.

17)Data Sufficiency->If M and N are positive integers greater than 1, what is the remainder when the expression \(22^{3M} * 39^{2N} + 14^{2(M+N)}\) is divided by 5? (1) M = 13 (2) N = 14

25)Data Sufficiency->Find the units digit of \(5^{3n} + 9^{5m}\) , where m and n are positive integers. (1)m is an odd integer. (2)n is an even integer.

26)If a is a positive integer, and if the units digit of \(a^2\) is 9 and the units digit of \((a+1)^2\) is 4, what is the units digit of \((a+2)^2\)? A)1 B)3 C)5 D)6 C)14

E. We are asked if n is an even integer or not. Statement 1-> n^2=n => n can be 0 or 1 0=even 1=odd Hence not sufficient. Statement 2-> n^3=n n=0 or 1 or -1 Hence not sufficient. Combing the two statements => n can be zero or one. zero is even one is odd.

Hence not sufficient.

Hence E.

66)Data Sufficiency->If n is an integer, is n even? (1) n^2 - 1 is an odd integer. (2) 3n + 4 is an even integer.

68)Data Sufficiency->If n and m are positive integers, is m a factor of n? (1) n = 5(3^k), for any positive integer k. (2) m = 3^(k-1), for any positive integer k.

Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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stonecold wrote:

If the GCD of m and 25 is 5 and GCD of m and 12 is 3,which of these can never be the value of positive integer m?

A)15 B)45 C)60 D)105 E)165

Test Using options -

A. \(15 = 3*5\), So, GCD ( 15, 25) is 5 & GCD ( 15, 12) is 3 B. \(45 = 3^2*5\), So, GCD ( 45, 25) is 5 & GCD ( 45, 12) is 3 C. \(60 = 2^2*3*5\), So, GCD ( 60, 25) is 5 & GCD ( 25, 12) is 1

We, need to go no further... Answer will be (C).. A quick check will be Using the options. _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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Word Problem Set based on Fractions, Ratio,Interest and Percents -->

Q1)Class B has 50% more students than class A. Number of girls in class A is equal to number of boys in class B. The percentage of girls is the same in both classes. What percentage of the student group are boys? A. 25 % B. 33 % C. 40 % D. 50 % E. 60 %

Q2)The Bryant's flower shop situated at the New Plaza complex stocks four types of flowers. There are 1/3 as many violets as carnations, and 1/2 as many tulips as violets. If there are equal no. of roses and tulips, what percent of the flowers in the shop are carnations?

Q4)XYZ Corporation has a ratio of 1 : 6 female to male. If the average salary for the female employees is $217,800 and XYZ pays $2,395,800 in salary to female employees, how many men work at XYZ Corporation?

Q5)Two libraries are planning to combine a portion of their collections in one new space. One third of the books from Library A will be housed in the new space along with One fourth of the books from Library B. If there are twice as man books in Library B as in Library A, what proportion of the books in the new space will have come from Library A?

Q6)Question 7) Lisa spends 3/8th of her salary on rent and 5/12 on food. Her roommate, Carrie earns about twice as much as Lisa, spends 1/4th of her salary on the rent and 1/2 on food. If the two women decide to contribute the rest of their salary to charity every month, what fraction of their combined monthly income will they donate.

No options for this one, just tell me the fraction value

Q7)Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec B. 14 min 52 sec C. 14 min 58 sec D. 16 min 6 sec E. 16 min 12 sec

Q8)Question 8) A restaurant spends one quarter of its monthly budget for rent and half of the rest for food and beverages. What percentage of the budget does the restaurant spend for food and beverages?

Q10)The ratio of a to b is 4 to 5, where a and b are positive. If x equals a increased by 25 percent of a, and m equals b decreased by 20 percent of b, what is the value of m/x?

Q13)Each month, after Jill pays for rent, utilities, food, and other necessary expenses, she has one fifth of her net monthly salary left as discretionary income. Of this discretionary income, she puts 30% into a vacation fund, 20% into savings, and spends 35% on eating out and socializing. This leaves her with $96 dollar, which she typically uses for gifts and charitable causes. What is Jill’s net monthly salary?

Q14)At a speed of 50 miles per hour, a certain car uses 1 gallon of gasoline every 30 miles. If the car starts with a full 12 gallon tank of gasoline and travels for 5 hours at 50 miles per hour, the amount of gasoline used would be what fraction of a full tank?

Q15)At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?

Q16)At a certain school, the ratio of the number of second graders to the number of fourth graders is 8 to 5, and the ratio of the number of first graders to the number of second graders is 3 to 4. If the ratio of the number of third graders to the number of fourth graders is 3 to 2, what is the ratio of the number of first graders to the number of third graders?

A. 16 to 15 B. 9 to 5 C. 5 to 16 D. 5 to 4 E. 4 to 5

Q18)Initially, the men and women in a room were in the ratio of 4 : 5. Then, 2 men entered the room and 3 women left the room. Then, the number of women doubled. Now there are 14 men in the room. How many women are currently in the room?

Q19)A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?

Q20)Kumail, a noted sneaker enthusiast, has a collection consisting of sneakers made by Brand A and Brand B. 5/7 of the Brand A sneakers are low-tops, and 1/3 of the Brand A low-tops are running shoes. If Kumail owns 7 more pairs of Brand B sneakers than Brand A sneakers, what is the minimum possible number of pairs of sneakers in his collection?

Q21)If x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $56. If $2,000 is the total amount invested, how much is invested at 8 percent?

Q22) Last year Elaine spent 20% of her annual earnings on rent. This year she earned 15% more than last year and she spent 30% of her annual earnings on rent. The amount she spent on rent thisyear is what percent of the amount spent on rent last year?

Q23)On the first of the year, James invested x dollars at Proudstar bank in an account that yields 2% in interest every quarter year. At the end of the year, during which he made no additional deposits or withdrawals, he had y dollars in the account. If James had invested the same amount in an account which pays interest on a yearly basis, what must the interest rate be for James to have y dollars at the end of the year?

Set X of has an average of 61. If the largest element is 7 greater than 6 times the smallest element, how many values out of {2,3,5,8,9,11,53,102,123,178,210,267,283,311,376,383,399,401} can be a part of Set X?

Hi Since the numbers of elements in set X is not given, we can play around with the total numbers and what is important is the smallest and largest values possible..

Smallest:- Let the largest value be just above average.. So 6y+7>61....6y>61-7.....y>9

Largest:- Let the smallest value be just below 61.... So 6*61+7>y.......373>y..

Range becomes 9>y>373.. Values in this range are 11,53,102,123,178,210,267,283,311.....9 values Ans C
_________________

Its not any specific method. I just listed out the possible powers in two columns. Doing it will help us know that each value in column A will have 3 values in column B or vice versa.

Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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10 Jan 2017, 18:06

Here is my solution to this one => 9919 => 10000-81=> 100^2-9^2=> 109*91 => 109*7*13 => Thus the greatest prime factor =109 NOTE=> 109 is a prime as it is not divisible by 2,3,5,7 In order to check whether a number is prime or not we just need to divide it by all the primes less than equal to the square root of that number.

Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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stonecold wrote:

If N is a positive 3-digit number, is N prime?

(1)The hundreds digit of K is the sum of the tens and units digit. Tens and units digits are equal. (2)N is odd.

Source->Self-Made

stonecold There is a typo in Statement 1 . you have mentioned K instead of N .

On the Solution part . From Statement 1 , we know that N can be equal to 211, 422, 633, and 844 . As a note: Only 211 is a prime here . However , we don't get a unique value. So , Statement 1 is not Sufficient

Statement 2, Says N is odd N can be 123 (Not prime) or 131 ( Prime ) . No unique answer . So , Statement 2 is not Sufficient

Combine . N can be equal to 211 or 633 No unique value .

How many divisors does the positive integer \(N\) have?

A) \(27N^3\) has 16 factors. B) \(90<N^3<200\)

Source->Self-Made

Hi....

Let's see the statements..

A) \(27N^3\) has 16 factors. \(3^3*N^3\) has (3+1)(3+1)=16 factors... The number of factors are always taken by getting the number into prime factors. So here both 3 and N are PRIME numbers.. So N will have only two factors-- 1 and N.. Suff

B) \(90<N^3<200\) Only N as 5 fits in.. 4^3=64 and 6^3=216.. Sufficient

Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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stonecold wrote:

How many divisors does the positive integer \(N\) have?

A) \(27N^3\) has 16 factors. B) \(90<N^3<200\)

Source->Self-Made

From Statement 1 . Given \(27N^3\) has 16 factors => \(3^3 N^3\) has 16 factors Total factors are calculated by adding +1 to prime exponents and multiplying . In this case (3+1)(3+1) = 16 (3+1) is coming from 3 and another (3+1) need to come from N N must be a prime factor and can't be a composite number because we will get more multipliers and factors will be greater than 16 . As N is a prime number . We know it has only 2 factors .

Statement 1 is sufficient .

From Statement 2 . Only 5^3 fits . N= 5 Statement 2 is sufficient .

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